Question

# $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - \cos 2x)}^2}}}{{2x\tan x - x\tan 2x}}$ A) $2$ B) $- \dfrac{1}{2}$ C) $- 2$ D) $\dfrac{1}{2}$

Hint:In this question in numerator use, $\cos 2x = 1 - 2{\sin ^2}x$ , Afterwards multiple and divide by ${x^3}$ and we know that $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$ while in denominator use the expansion of $\tan x$ and simplify it to get required answer.

As in the given question $\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - \cos 2x)}^2}}}{{2x\tan x - x\tan 2x}}$ first in the numerator
Use the formula $\cos 2x = 1 - 2{\sin ^2}x$
Hence
$\mathop {\lim }\limits_{x \to 0} \dfrac{{{{(1 - 1 + 2{{\sin }^2}x)}^2}}}{{2x\tan x - x\tan 2x}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4{{\sin }^4}x}}{{x(2\tan x - \tan 2x)}}$
Now multiple and divide by ${x^3}$ than use the property $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4{{\sin }^4}x \times {x^3}}}{{{x^4}(2\tan x - \tan 2x)}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^3}}}{{(2\tan x - \tan 2x)}}{\left( {\dfrac{{\sin x}}{x}} \right)^4}$
As from above $\mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1$
Then $\mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^3}}}{{(2\tan x - \tan 2x)}}$
Now for the denominator we have to use the expansion of $\tan x$
We know that the expansion of $\tan x = x + \dfrac{{{x^3}}}{3} + \dfrac{{2{x^5}}}{{15}}..........$
Similarly $\tan 2x = 2x + \dfrac{{{{(2x)}^3}}}{3} + \dfrac{{2{{(2x)}^5}}}{{15}}..........$
Now by putting these values in the above equation
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^3}}}{{\left\{ {2\left( {x + \dfrac{{{x^3}}}{3} + \dfrac{{2{x^5}}}{{15}}..........} \right) - \left( {2x + \dfrac{{{{(2x)}^3}}}{3} + \dfrac{{2{{(2x)}^5}}}{{15}}..........} \right)} \right\}}}$
$\mathop {\lim }\limits_{x \to 0} \dfrac{{4{x^3}}}{{\left( {2x + \dfrac{{2{x^3}}}{3} + \dfrac{{4{x^5}}}{{15}}.......... - 2x - \dfrac{{{{(2x)}^3}}}{3} - \dfrac{{2{{(2x)}^5}}}{{15}}........} \right)}}$
On dividing by ${x^3}$ on denominator and numerator

$\mathop {\lim }\limits_{x \to 0} \dfrac{4}{{\left( {\dfrac{2}{3} + \dfrac{{4{x^2}}}{{15}}.......... - \dfrac{8}{3} - \dfrac{{2{{(2)}^5}{x^2}}}{{15}}........} \right)}}$
By applying $\mathop {\lim }\limits_{x \to 0}$ , Now all the term containing x will equal to $0$
$\dfrac{4}{{\left( {\dfrac{2}{3} - \dfrac{8}{3}} \right)}}$
$\dfrac{4}{{ - 2}}$
That is equal to $- 2$.

So, the correct answer is “Option C”.

Note:Whenever the trigonometric functions are not solved further in limit problems apply the expansion of the trigonometric function then it will solve easily without taking much time .Always expand the limit the power of x in the numerator is equal or greater to the power of the denominator otherwise the question will not proceed to the answer .